X dived by 3 what is it
2 answers:
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Answer:
1.314 MJ
Step-by-step explanation:
As water is removed from the tank, decreasing amounts are raised increasing distances. The total work done is the integral of the work done to raise an incremental volume to the required height.
There are a couple of ways this can be figured. The "easy way" involves prior knowledge of the location of the center of mass of a cone. Effectively, the work required is that necessary to raise the mass from the height of its center to the height of the discharge pipe.
The "hard way" is to write an expression for the work done to raise an incremental volume, then integrate that over the entire volume. Perhaps this is the method expected in a Calculus class.
<h3>Mass of water</h3>The mass of the water being raised is the product of the volume of the cone and the density of water.
The cone volume is ...
V = 1/3πr²h . . . . . . for radius 2 m and height 8 m
V = 1/3π(2 m)²(8 m) = 32π/3 m³
The mass of water in the cone is then ...
M = density × volume
M = (1000 kg/m³)(32π/3 m³) ≈ 3.3510×10^4 kg
<h3>Center of mass</h3>The center of mass of a cone is 1/4 of the distance from the base to the point. In this cone, it is (1/4)(8 m) = 2 m from the base.
<h3>Easy Way</h3>The discharge pipe is 2 m above the base of the cone, so is 4 m above the center of mass. The work required to lift the mass from its center to a height of 4 m above its center is ...
W = Fd = (9.8 m/s²)(3.3510×10^4 kg)(4 m) = 1.3136×10^6 J
<h3>Hard Way</h3>
As the water level in the conical tank decreases, the remaining volume occupies a space that is similar to the entire cone. The scale factor is the ratio of water depth to the height of the tank: (y/8). The remaining volume is the total volume multiplied by the cube of the scale factor.
V(y) = (32π/3)(y/8)³
The differential volume at height y is the derivative of this:
dV = π/16y²
The work done to raise this volume of water to a height of 10 m is ...
(9.8 m/s²)(1000 kg/m³)(dV)((10 -y) m) = 612.5π(y²)(10 -y) J
The total work done is the integral over all heights:
It takes about 1.31 MJ of work to empty the tank.
Answer:
Part A) The length of a straight line between the school and the Fire Station is
Part B) The length of a straight line between the school and the hospital is
Step-by-step explanation:
<u><em>The complete question in the attached figure</em></u>
Let
The positive x-coordinates ----> East
The positive y-coordinates ----> North
The negative x-coordinates ----> West
The negative y-coordinates ----> South
take the point A (0,0) as the Middle School (reference point)
Part A) we have
Town Hall is located 4.3 miles directly east of the Middle School
so
The coordinates of Town Hall are B(4.3,0)
The Fire Station is located 1.7 miles directly North of Town Hall
so
The coordinates of Fire Station are C(4.3,1.7)
What is the length of a straight line between the school and the Fire Station?
Remember that
the formula to calculate the distance between two points is equal to
we have
A(0,0) and C(4.3,1.7)
substitute
Part B) we have that
The hospital is 3.1 miles west of the fire station
so
The coordinates of the hospital are D(4.3-3.1,1.7)
D(1.2,1.7)
What is the length of a straight line between the school and the hospital?
Remember that
the formula to calculate the distance between two points is equal to
we have
A(0,0) and D(1.2,1.7)
substitute
